## Calculus: Early Transcendentals 8th Edition

$e^{nx}$
$(coshx+sinhx)^{n}$ = cosh nx + sinh nx Convert the left side to the $e^{x}$ forms and simplify $(\frac{e^{x} + e^{-x}}{2} + \frac{e^{x} - e^{-x}}{2} )^{n}$ = $(\frac{2e^{x}}{2})^{n}$ = $(e^{x})^{n}$ = $e^{nx}$